lecture06

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S. Yan, EE 338L Lecture 6 1 EE 338L CMOS Analog Integrated Circuit Design Lecture 6, Single-Stage Amplifiers (3) Cascode Amplifiers We will cover different cascode amplifiers, including 1. Simple cascode amplifier 2. Multi-level cascode amplifier 3. Gain boosted cascode amplfier 4. Folded cascode amplifier 1. Simple cascode amplifier M1 M2 v out v in Vdd I B V B Large signal behavior (Vin fixed to V G1 , Vout (V DS ) sweeping from 0 to 3V) I D,CASCODE I D,SIMPLE I D V DS I II III V G1 M1A I D,SIMPLE V G2 M1B M2B V G1 I D,CASCODE Region I: M1B and M2B both in triode; Region II, M1B in saturation, M2B in triode; Region III, M1B and M2B both in saturation

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S. Yan, EE 338L Lecture 6 2 Small signal analysis We will calculate small signal i) output resistance, ii) transconductance (when output is shorted to a fixed DC voltage), iii) DC voltage gain (when the output is open). i) Output resistance We have derived earlier, [] 2 1 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 ) 1 ( ) 1 ( ) ( ds ds ds m ds ds ds ds m s ds ds ds ds mb m out r r r g r r r r g r r r r g g r + + + = + + + = + + + = η ii) Transconductance (when output is shorted to a fixed DC voltage or AC ground) Short the output to an AC ground, and draw the small signal diagram as shown above. According to KCL we can list the following equations, 12 11 23 22 21 i i i i i i out + = + + = (1) where in m gs m v g v g i 1 1 1 11 = = ( 2 a ) 2 1 12 s ds v g i = ( 2 b ) v g2 v in g ds2 v out =0 i 11 =g m1 v gs1 = g m1 v in g ds1 v gs2 v gs1 v s2 v b2 g mb1 v bs1 =g mb1 0=0 v b1 v bs2 v bs1 =0 i 21 =g m2 v gs2 i 23 =g mb2 v bs2 i 22 i out i 12 2 2 2 2 2 2 2 21 ) ( s m s g m gs m v g v v g v g i = = = (2c)
S. Yan, EE 338L Lecture 6 3 2 2 2 2 2 2 2 22 ) ( s ds s d ds ds ds v g v v g v g i = = = (2d) 2 2 2 2 2 2 2 23 ) ( s mb s b mb bs mb v g v v g v g i = = = (2e) From Eq. (1), we have ) ( 23 22 21 i i i i out + + = ( 3 a ) Substitute Eqs. (2c)-(2e) into Eq. (3a), we have, 2 2 2 2 23 22 21 ) ( ) ( s ds mb m out v g g g i i i i + + = + + = (3b) From Eq. (1), we have 12 11 23 22 21 i i i i i + = + + ( 4 a ) Substitute Eqs. (2a)-(2e) into Eq. (4a), 2 1 1 2 2 2 2 ) ( s ds in m s ds mb m v g v g v g g g + = + + (4b) Solving Eq. (4b), we get in ds ds mb m m s v g g g g g v 1 2 2 2 1 2 + + + = (5) Substitute Eq. (5) into Eq. (3b), in ds ds mb m ds m in ds ds mb m ds mb m m s ds mb m out v g g g g g g v g g g g g g g g v g g g i + + + = + + + + + = + + = 1 2 2 2 1 1 1 2 2 2 2 2 2 1 2 2 2 2 1 ) ( (6) Thus the transconductance of the cascode amplifier is 1 1 2 2 2 1 1 1 2 2 2 2 2 2 1 1 m ds ds mb m ds m ds ds mb m ds mb m m in out m g g g g g g g g g g g g g g g v i G + + + = + + + + + = = (7) Observation: Compared with a single-transistor common source amplifier with a transconductance of | G m |= g m1 (Note that G m is the transcoducance of the amplifier , and g m is the transcoducance of the transistor ), the transcoductance of cascode amplifier is slightly less, whose transconductance Gm is given by 1 1 2 2 2 1 1 99%) to % 90 ( 1 m ds ds mb m ds m m g g g g g g g G = + + + = .

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S. Yan, EE 338L Lecture 6 4 iii) DC voltage gain (when the output is open) v g2 v in g ds2 i 11 =g m1 v gs1 = g m1 v in g ds1 v gs2 v gs1 v s2 v
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## This note was uploaded on 10/23/2011 for the course SDASD 102 taught by Professor Dsfas during the Spring '11 term at Baptist Bible PA.

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